# Differentiating Logarithmic Functions with Base e

Derivatives #8: Logarithmic Differentiation

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Darshan Senthil

## Key Questions

• $\frac{1}{x}$ is the derivative of ln(x).

Let us derive the above derivative by Implicit Differentiation.

By rewriting in terms of an exponential function,

$y = \ln x R i g h t a r r o w {e}^{y} = x$

by implicitly differentiating with respect to $x$,

$R i g h t a r r o w {e}^{y} \cdot y ' = 1$

by dividing by ${e}^{y}$,

$R i g h t a r r o w y ' = \frac{1}{e} ^ y = \frac{1}{x}$.

Hence, $\left(\ln x\right) ' = \frac{1}{x}$.

I hope that this was helpful.

• $y ' = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{{x}^{2}} = \frac{1 - \ln x}{{x}^{2}}$

This problem can also be solved by the Product Rule

$y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g \left(x\right)$

The original function can also be rewritten using negative exponents.

$f \left(x\right) = \ln \frac{x}{x} = \ln \left(x\right) \cdot {x}^{-} 1$

$f ' \left(x\right) = \frac{1}{x} \cdot {x}^{-} 1 + \ln \left(x\right) \cdot - 1 {x}^{-} 2$

$f ' \left(x\right) = \frac{1}{x} \cdot \frac{1}{x} + \ln \left(x\right) \cdot - \frac{1}{x} ^ 2$

$f ' \left(x\right) = \frac{1}{x} ^ 2 - \ln \frac{x}{x} ^ 2$

$f ' \left(x\right) = \frac{1 - \ln \left(x\right)}{x} ^ 2$

• The answer would be $f ' \left(x\right) = \frac{1}{g} \left(x\right) \cdot g ' \left(x\right)$ or it can be written as $f ' \left(x\right) = \frac{g ' \left(x\right)}{g} \left(x\right)$.

To solve this derivative you will need to follow the chain rule which states:

$F \left(x\right) = f \left(g \left(x\right)\right)$ then $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.

The derivative of $h \left(x\right) = \ln \left(x\right)$ is $h ' \left(x\right) = \frac{1}{x}$.

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